Having trouble setting up and evaluating the following problem:
If a freely falling body starts from rest, then its displacement is given by s=1/2gt^2. Let the velocity after a time T be v_T. Show that if we compute the average of the velocities with respect to t we get v_ave = 1/2v_T, but if we compute the average velocities with respect to s we get v_ave = 2/3v_T
How would I solve this?
So we have $s(t) = \frac{1}{2}gt^2$ therefore $v(t) = gt$. So compute the average velocity with respect to time, then we integrate the velocity over time and divide by the total time:
$$ v_{avg_t} = \frac{\int\limits_0^T gtdt}{T - 0} = \frac{\frac{1}{2}gT^2}{T} = \frac{1}{2}gT = \frac{1}{2}v_T $$
because $v(T) = gT = v_T$. BTW, this is the usual definition of "average velocity": $v_{avg} = \frac{\Delta s}{\Delta t}$.
If, however, we want to find the spacial average, that is the average velocity over the displacement, then we need to do a different integral:
$$ v_{avg_s} = \frac{\int\limits_{s_0}^{s_T}v(t)ds}{s_T - s_0} $$
Notice that we have a mix of variables: $t$ and $s$. We can use substitution to find $ds$ in terms of $dt$:
$$ \frac{ds}{dt} = v(t) \rightarrow ds = v(t)dt $$
Therefore we have:
$$ v_{avg_s} = \frac{\int\limits_{0}^{T}v^2(t)dt}{s_T - s_0} = \frac{\int\limits_{0}^{T}g^2t^2dt}{s_T - s_0} = \frac{\frac{g^2T^3}{3}}{\frac{1}{2}gT^2} = \frac{2}{3}gT = \frac{2}{3}v_T $$
Notice that in the substitution the limits of integration "magically" changed from $s_0$ to $s_T$ to $t = 0$ to $t = T$. This is because, since we changed the integration to be over $t$ we need to find the $t$ values where $s = s_0$ and $s = s_T$--by definition these must occur at $t = 0$ and $t = T$, respectively.
Alternatively, you could find $v$ in terms of $s$:
\begin{align} s(t) = \frac{1}{2}gt^2 \rightarrow&\ t = \sqrt{\frac{2s}{g}}\\ v(t) = gt \rightarrow&\ v(s) = g\sqrt{\frac{2s}{g}} = \sqrt{2gs} \end{align}
Now the integral becomes
$$ \require{cancel} \sqrt{2g}\int\limits_0^{\frac{1}{2}gT^2} \sqrt{s}ds = \sqrt{2g}\frac{2}{3}\left(\frac{1}{2}gT^2\right)^{\frac{3}{2}} = \frac{1}{3}g^{\frac{1}{2}}g^{\frac{3}{2}}\frac{\cancel{2^{\frac{1}{2}}\cdot 2}}{\cancel{2^{\frac{3}{2}}}}T^3 = \frac{g^2T^3}{3} $$
which is the same result we got from the substitution above.